I recently learned that if you have a centrifuge whose number of holes $n$ is divisible by $6$, then you can balance any number of tubes except for $1$ and $n-1$. If $k$, the number of tubes you want to balance, is even, just put them in $k/2$ pairs of opposite holes. If $k$ is odd and bigger than $1$ and less than $n-1$, then you can place $3$ tubes in an equilateral triangle, leaving $n/2-3$ unoccupied pairs of holes for the remaining $(k-1)/2$ tubes. I was wondering if we could solve the problem for a general $n$.
To make this more formal, let $\zeta_n$ be a primitive $n$th root of unity. We can balance $k$ tubes in a centrifuge of $n$ holes if there exists a subset of $\{1, \zeta_n,\ldots, \zeta_n^{n-1}\}$ whose sum is $0$.
If $n=p$ is prime, then you can only balance $0$ or $p$ tubes, since given a nonempty proper subset of the $p$th root of unity WLOG we can choose $\zeta_p$ so that $\zeta_p^{p-1}$ is not in the subset (unless the subset is everything except $1$, which clearly doesn't sum to $0$), and the zero sum gives a nonzero polynomial in $\zeta_p$ of degree less than $p-1$, contradicting $1+x+\cdots+x^{p-1}$ being the minimal polynomial of $\zeta_p$.
I've made a few more observations: in general, if $k$ and $n$ have a common factor greater than $1$, then you can balance $k$ tubes in a centrifuge with $n$ holes (just take the disjoint union of $n/k$ cyclically symmetric $k$-tuples). I suspect this is the only possibility if $n=p^m$ is a prime power (i.e., $k$ must be divisible by $p$). Furthermore, I suspect that if $n=pq$ is the product of distinct primes, then only multiples of $p$ and multiples of $q$ work: I couldn't do $k=p+q$ as a disjoint union, since if you take the $q$ roots whose powers are the multiples of $p$, then each is unique mod $q$, so any cyclically symmetric $p$-tuple will intersect with the $q$-tuple.
In general, I suspect that the only way to balance tubes is to start with cyclically symmetric arrangements and take successive disjoint unions and set differences. Has this been shown?